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Conservation of momentum and lost energy

  • Thread starter RubinLicht
  • Start date
1. The problem statement, all variables and given/known data
sorry for the long question.

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2. Relevant equations


3. The attempt at a solution
I get everything up until it asks where the extra "half" of the power is spent. I do know that the extra half is referring to the "2" dK/dt. I have spent the last hour or so thinking about this while working the chapter practice problems out, just in case i glimpse a moment of realization by thinking about energy. Unfortunately no.

Since this is not a "homework problem" could someone just explain why it is 2dK/dt and not the more logical dK/dt?
Energy lost during collision of sand with belt? perhaps something to do with the fact that the belt is a belt and not a plane moving perpetually right?
 

TSny

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Think of one of the grains of sand as a little block that is dropped onto the moving belt. Think about how it looks in a reference frame moving with the belt. In this frame the belt is at rest and the block has an initial horizontal speed vo just before it lands on the belt. Then the block skids to rest on the belt. So, in this frame of reference, all the initial kinetic energy of the block (½mvo2) is "lost". What actually happened to that energy?
 
Think of one of the grains of sand as a little block that is dropped onto the moving belt. Think about how it looks in a reference frame moving with the belt. In this frame the belt is at rest and the block has an initial horizontal speed vo just before it lands on the belt. Then the block skids to rest on the belt. So, in this frame of reference, all the initial kinetic energy of the block (½mvo2) is "lost". What actually happened to that energy?

Oh yea I recall thinking about where the ke went but I just skipped over that thought... Oops. Thanks tho
 
I came across this problem and found it pretty interesting .

Consider the lab frame and (hopper+sand+belt) as the system .

Power input by external force acting on the belt = ΔK/Δt ( rate of change of KE of falling sand ) + ΔE/Δt ( rate of generation of thermal energy due to friction between falling sand and belt ) .

Now , switch to a frame moving with the belt and sand falling on the belt as the system .

Applying Work KE theorem , W ( work done by friction ) = ΔK ( change in KE of the falling sand ) . Further , W = ΔE as work done by friction is dissipated as thermal energy .

So , ΔK = ΔE and P = 2ΔK/Δt .

@TSny , Is there any flaw in this reasoning ?
 
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TSny

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Is there any flaw in this reasoning ?
I think it's a good argument. The "work done by friction" can be tricky (see http://www4.ncsu.edu/~basherwo/docs/Friction1984.pdf). But the way you used this concept looks ok to me. The important thing in your argument is that in the belt frame, the initial kinetic energy of the sand is completely converted into thermal energy.

Also, you are switching between frames in your analysis. How would you answer someone who points out that the change in kinetic energy of an object as measured in one frame is generally different than the change in kinetic energy of the object as measured in a different frame? Does ΔKsand in the belt frame equal ΔKsand in the lab frame?
 
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Also, you are switching between frames in your analysis. How would you answer someone who points out that the change in kinetic energy of an object as measured in one frame is generally different than the change in kinetic energy of the object as measured in a different frame?
Fair enough :smile:

Does ΔKsand in the belt frame equal ΔKsand in the lab frame?
Yes . In the belt frame , a grain of sand decelerates from -v to 0 , ΔK = ½mv2

In the lab frame , a grain of sand accelerates from 0 to v , ΔK = ½mv2 .

We can see the change in KE in both the frames is equal .
 

TSny

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OK, but note that ΔK has opposite sign for the two frames of reference.
 

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